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1、The second edition of electromagnetism is the answer to the first chapterThe basic law of static fieldTo determine whether the following statement is correct and justifications.(1) a little bit of field intensity is the direction of the electric field force of the probe point charge.(2) the intensit
2、y of the field can be determined by E = F/q, where q is positive and negative.(3) on a sphere that is centered on a point charge, the field intensity generated by this point charge is the same everywhere.(1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1 (2) the square root; (3) x is on the surface of the s
3、phereE?Equal size.How do you use symmetry to determine the direction of the center of the ball?Answer: using the symmetry analysis, the components of the vertical axis cancel each other.Is the following statement true? Why is that?(1) the total charge must be zero for each point on the closed surfac
4、e at zero.(2) when the total charge in the closed surface is zero, each point on the surface will be zero.(3) when the E flux of the closed surface is zero, each point on the surface will be zero.(4) the E flux on the closed surface is provided only by the surface charge.(5) the intensity of each po
5、int on the closed surface is only provided by the internal charge.(6) the condition of the gauss theorem is symmetrical.(7) the field intensity obtained by gauss theorem is only stimulated by the charge of charge in gauss.(1) no net charge; (2) x; (3) x; (4); (5) x; (6) x; (7) x.1.4 the field intens
6、ity of a uniform charged sphere is equal to the intensity of the field at the center of the ball.Really?The answer: in the absence of the field, it is correct for the ball.A and B are two homogeneous electric bodies, S for the sphere of A.(1) does the flux in S surface have to do with the position o
7、f B and the charge?(2) does the strength of an electric field at some point on the S surface have to do with the position of B and the charge?(3) can you use gausss theorem to find a point on the surface of S? Why is that?(1) it doesnt matter (2) it doesnt matter (a conductor ball).When the principl
8、e of field superposition is applied to a problem with a conductor, it is important to note that when a charged conductor exists alone, there is a charge distributionChange the pageThey produce an electric field; When n live conductors are put together, because of electrostatic induction, the charge
9、distribution on the conductor changes, and then,The principle of applied superposition should freeze the charge distribution of different conductors, and then store it separately with the frozen charge distributionThe electric field generated by the time is superimposed.The radius Rs army is able to
10、 dig a small ball with radius R inside the ball, and can use gauss to determine the two kinds of dredging (a) and (b) of the attached image (a) and (b)How does the principle of theorem and superposition work?The answer: (a), the superposition method (compensatory method); (b) notS1, S2, S3, and S4 i
11、n the attached chart are all curved surfaces with closed curve L (the direction of the surface normal line). Have been S1E flux forone, surface S2, S3 and S4 E flux2,3 and4.The point E = 0 constant, always outside204qER PI epsilon.= constant,The point where the surface of the balloon is passing by,
12、the E is going to change2004qER PI epsilon.= .S1 and S2 are four closed surfaces in the attached chart, E1, E2, and E3 respectively represent the static field strength of q1, q2, q3, and try to judgeThe right and wrong of the following equations(1)oneoneone0s.qEdsEpsilon.? = ?(2)2330s.qEdsEpsilon.?
13、= ?(3)one2320(a)s.qEEdsEpsilon.+? = ? ?(4)one12120(a)(a)s.qqEEdsEpsilon.+? = ? ?(5)2321230(a)(a)s.qqEEEdsEpsilon.+ +? = ? ?(6)one1321230(a)(a)s.: QQQEEEdsEpsilon.+ + +? = ? ?Answer: (1) x; (2) x; (3) x; (4); (5); (6) x;The electric field diagram of the two infinite uniformly charged plane with the e
14、quivalent number and the equivalent number is drawn.The answer:Does the 1.10 electric field line be the trajectory of the electric charge in the field? (set up thisChange the pageThe point charge is not subject to any other force except the electric field.Answer: not usually. FqE = ; FMa = ;va.t=?;
15、Only in the uniformly strong electric field, the orbit of the stationary point charge motionThe power line of the trace.Are the following statements true? If incorrect, please give a counter-example.(1) the potential of the field is equally small.(2) if two electric potentials are equal, their field
16、 strength is equal.(3) if A point is stronger than B, it is bound to be higher than B.(4) the field strength must be zero.(5) the potential for zero is zero.Answer: (1) incorrect.uEnn?=?For example, the uniform electric field.(2) incorrect.(3) incorrect. E big, the rate of change of electric potenti
17、al is big, not necessarily U big(4) incorrect. E = 0,Un?Its 0, and its not U must be 0, in the same amount as the point of the same charge.(5) incorrect. U = 0, is notUn?Must be 0, for example: in the same amount of different points of charge.The 1.12 radius is R1 and R2 = 2R1, which is the concentr
18、ic uniform sphere with charge q10. When the outer sphere has charge q2What is the potential for the inner sphere to be positive? What are the conditions for which the inner sphere is zero? What is the negative of the inner sphere?(reference points are not far away.)The answer:12one0101442qqUThe RR P
19、I epsilon PI epsilon.= +Or:21121121121122220044The RRRRRR: QQQUEdrEdrdrdrThe rr PI epsilon PI epsilon.upUp += = + + To make the10U,21 () 02qQ +, that is,212Qq?To make the10U =,21 () 02qQ + =, that is,212Qq =?To make the10U,21 () 02qQ +, that is,212qqThe field intensity of the area of the area of the
20、 area of the area of the area of the area of the area of the area of the area of the area of the area of the area of the area of the area of the areaProof: if E is 0, 0baba.UEdl = = , that is,abUU =, the allelic region.If its an allele, U = 0, thats 0UEn?= =?It is.The large round cross section of th
21、e uniform charged hemisphere surface (see attached) is equal to the potential surface. (hint: make up the other half of the sphere, for symmetryChange the pageThe field strength of each sphere on S is perpendicular to S.Prove that there is a field force parallel to the component of the s surface, an
22、d the total field strength of the change point in the ball after the other half of the ballIt should be zero, so you cant have a strong parallel component on the s surface, and its only going to be the vertical component of the fieldThe s is equal to the potential surface.The 1.2.1 vacuum has two po
23、int charges, one of which is four times the value of the other. They are separated by2 -So 5.010 times m is repulsive1.6 N. Q:(1) what is their charge?(2) how much are they repulsive when they are 0.1 m apart?Solution: set a chargeoneQ,214Qq = by the formula1220one4qqFR PI epsilon.= you can get:2one
24、220411.64 (510).qPI epsilon.?=xSolution of:oneQ =60.3310? x2,14Qq = =61.3310? x when r = 0.1, by repulsion is:1220one4 (0.1)qqFPI epsilon.= = 0.4 (N)1.The sum of charge of the charge of 2.2 and two of the same charge is Q, at which point the charge is equal to each otherThe maximum interaction?Solut
25、ion: lets say one charge is q, and the other charge is q-q,By the coulomb force2(a) : qQqFk1 - fkr?= knowWhen 0dfdq= that is:220K.Qqr? =So the power of the two is: 2Qq =one2QQ =The sum of the charge of the charged body of a point charged at 1.2.3 is 2q and q, respectivelyForce is zero?Solution: set
26、a qDistance q r, q)2 q (Lr)? In the distance r, the force is equal to zero.222(a)QQQQkkrLr =?Change the pageGet: (21) rL =?1.2.4 in rectangular coordinates (0, 0.1 m) and (0, -0.1 m); Two positions have charge q = 10- 10 cThe point ofThe charged body is placed on the (0.2 m, m) position with a charg
27、e of Q = 10- 8 cThe point of the charge, the magnitude and direction of the force of Q.According to the diagram, you can get:722220.2(23.2210)0.10.20.10.2QqfkN?= = x+i.Direction: horizontal to right1.2.5 each place an electric charge on the vertices of a square.(1) the force applied to any point cha
28、rge placed in a square center is zero.(2) if you put a point of charge at the center, make every charge at the vertex a reasonable amount of zero, the relationship between Q and Q.Answer: (1) the symmetry is known: O point E = 0, and the force charged at O point for any charge is 0.(2) the solution:
29、 set O point for a point of charge. 1220one4qQffA PI epsilon.= =(a)2320one42qfa.PI epsilon.=20one4one22qQFa.PI epsilon.=?To make the net force of q zero, we have:220222000121cos454442one22qQqqaaa.PI epsilon PI epsilon PI epsilon= +?Solution of:12(a)42Qq = + (Q should be negative point charge)The iso
30、tropic point charge of 1.2.6 is equal to 2a, and a pilot charge q0 is placed on the central vertical surface of the two lines, and q0 is the most stressedThe trajectory of a large point.Solution: 3222222222(a)(a)(a)QQRKQQRFk1 - fkararar = =+And 0dfDr.=Change the pageThat is:3122222222332 () ()220(a)
31、arrarrKQQar?+? X +?=?+?i.312222222() 3 () + =222Ar =So the radius of the circle is:22Ra = -So this is the equation of the curve:2222a.yz?+ =?. The spherical equation1.3.1 the electric field between the two parallel plates with the length of 50cm, which is 1cm apart, is the uniform electric field (ve
32、rtical upward in the field),The initial velocity of an electron from P to the top of the board is equal to 107 m/sThe level is fired into the field (see attached diagram). If the electron is on the bottomWere going to leave the field by the side, and were going to find the magnitude of the uniform f
33、ield. Ignore the edge effect, think that the outside field is zero, and take a little gravity to the electron transportThe effect of moving)Solution: the force of the electrons in the electric field produces motion acceleration:0EEma =From the kinematics equation:2122dYat = =0XVT = (2dY =) xL =Solut
34、ion of:20VMDEThe ex?=?1.3.2 a small ball with a fine line of 0.2 g, placed in a parallel-plate between two vertical plates (seeThe attached figure is the charge of the ball9 -610 x C,Want to make the fine line of hanging ball and electric field Angle is 600,Take the field between the two boards.Solu
35、tion: as shown in the figure:00Cos30cos60eEmg =Among them:0Cos60EqT =0Cos30mgT =Solution of:030mgEtge=1.3.3 electron ejectock intensity is3510 times N over C, the uniform electric field in the vertical direction, the initial velocity of the electron710 m/sAnd levelThe Angle of incidence is 300(see a
36、ttached figure), regardless of the influence of gravity, please:Change the page(1) the maximum height of the electron ascent.(2) the electron returns to its original height.Solution: (1) electronic force: fmaeE = =020onesin302Yvtat =? (1)00 sin30yvvThe at =?At maximum height: 0yv=the000Sin30vat =?00
37、 sin30vta.= (2)(1) the maximum height2000000sin30sin301sin302V. v.yvaaa?= x?220011Sin30 ()2vaa=?2200oneSin30 ()2va.=22002onesin30EemV =208The mvEe=(2) when returning to the horizontal position:Y = 0, namely:020onesin3002eEvtm? =Solution of:002sin30mvteE=So at the original height, the range is:20003c
38、os302The mvXVTeE= =The charges for the 1.3.4 electron were first measured by the mirigan experiment, which was designed by mirigan, as shown in the attached image.A small charged oil droplet in the electric field E, regulates E to balance the electric field force on the oil droplets with the weight
39、of the oil droplet, if the oil dropsThe radius of4 -1.6410 times cm, the equilibrium time E is equal to51.9210 x N/C. O:Change the page(1) the oil density is 0.851 g/cm3And the absolute value of the warm drop.(2) how many times does the value of this value charge e?Solution: (1) a little(2) mgqE =31948.02103mgRgqEEPI rho?Is equal to x coulomb1.3.5 points of charge q1 = 4.0 uc and q2 = 8.0 uc are 10cm apart, and each of them is 10cm field s